The foot of perpendicular of the point (2,0,5) on the line x+12=y-15=z+1-1 is (,,).

Question

The foot of perpendicular of the point $(2,0,5)$ on the line $\dfrac{x+1}{2}=\dfrac{y-1}{5}=\dfrac{z+1}{-1}$ is $(\alpha,\beta,\gamma)$. Which of the following is NOT correct?

Accepted Answer

$$
\mathrm{L}: \frac{\mathrm{x}+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}=\lambda \text { (let) }
$$

Let foot of perpendicular is

$$
\begin{aligned}
& \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \\
& \overrightarrow{\mathrm{PA}}=(3-2 \lambda) \hat{\mathrm{i}}-(5 \lambda+1) \hat{\mathrm{j}}+(6+\lambda) \hat{\mathrm{k}} \\
& \text { Direction ratio of line } \Rightarrow \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\
& \text { Now, } \Rightarrow \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{b}}=0 \\
& \Rightarrow 2(3-2 \lambda)-5(5 \lambda+1)-(6+\lambda)=0 \\
& \Rightarrow \lambda=\frac{-1}{6} \\
& \mathrm{P}(2 \lambda-1,5 \lambda+1,-\lambda-1) \equiv \mathrm{P}(\alpha, \beta, \gamma) \\
& \Rightarrow \alpha=2\left(-\frac{1}{6}\right)-1=-\frac{4}{3} \Rightarrow \alpha=-\frac{4}{3} \\
& \Rightarrow \beta=5\left(-\frac{1}{6}\right)+1=\frac{1}{6} \Rightarrow \beta=\frac{1}{6} \\
& \Rightarrow \gamma=-\lambda-1=\frac{1}{6}-1 \Rightarrow \gamma=-\frac{5}{6}
\end{aligned}
$$

Solutions · 1